The Self-Cultivation of a SQL Engineer
Since I personally prefer using window functions, I prioritized them in my solutions. Of course, these may not always be the optimal approaches—the answers are for reference only.
Updated:
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The Self-Cultivation of a SQL Engineer
https://leetcode-cn.com/problemset/database/
All the problems are from LeetCode; you can click on a problem to see the corresponding link.
Since I personally prefer using window functions, I've prioritized them in my solutions. Of course, these may not all be the optimal solutions; the answers are for reference only.
Each problem is followed by its difficulty level. If you don't have time to solve them all, you can practice by filtering problems on LeetCode based on their frequency of appearance.
I solved the problems in order and will continue to update this list.
175. Combine Two Tables
Difficulty: Easy
SQL Schema
Table 1: Person
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| PersonId | int |
| FirstName | varchar |
| LastName | varchar |
+-------------+---------+
PersonId is the primary key column for this table.Table 2: Address
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| AddressId | int |
| PersonId | int |
| City | varchar |
| State | varchar |
+-------------+---------+
AddressId is the primary key column for this table.Write a SQL query for a report that provides the following information for each person in the Person table, regardless if there is an address for each of those people:
FirstName, LastName, City, Stateselect FirstName,LastName,City,State
from Person p
left join Address a
on a.PersonId = p.PersonId176. Second Highest Salary
Difficulty: Easy
SQL Schema
Write a SQL query to get the second highest salary from the Employee table.
+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+For example, given the above Employee table, the query should return 200 as the second highest salary. If there is no second highest salary, then the query should return null.
+---------------------+
| SecondHighestSalary |
+---------------------+
| 200 |
+---------------------+SELECT
IFNULL(
(SELECT DISTINCT Salary
FROM Employee
ORDER BY Salary DESC
LIMIT 1 OFFSET 1),
NULL) AS SecondHighestSalary177. Nth Highest Salary
Difficulty: Medium
Write a SQL query to get the nth highest salary from the Employee table.
+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+For example, given the above Employee table, n = 2 should return the second highest salary 200. If there is no nth highest salary, then the query should return null.
+------------------------+
| getNthHighestSalary(2) |
+------------------------+
| 200 |
+------------------------+CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
RETURN (
SELECT IFNULL(
(select salary
from(
select salary,
rank() over(order by salary desc) rk
from Employee
group by salary
)t1
where rk=N),NULL) SecondHighestSalary
);
END178. Rank Scores
Difficulty: Medium
SQL Schema
Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no "holes" between ranks.
+----+-------+
| Id | Score |
+----+-------+
| 1 | 3.50 |
| 2 | 3.65 |
| 3 | 4.00 |
| 4 | 3.85 |
| 5 | 4.00 |
| 6 | 3.65 |
+----+-------+For example, given the above Scores table, your query should generate the following report (order by highest score):
+-------+------+
| Score | Rank |
+-------+------+
| 4.00 | 1 |
| 4.00 | 1 |
| 3.85 | 2 |
| 3.65 | 3 |
| 3.65 | 3 |
| 3.50 | 4 |
+-------+------+Important Note: For MySQL solutions, if you want to escape a reserved word used as a column name, you can use an apostrophe before and after the keyword. For example Rank
select Score,
dense_rank() over(order by Score desc) `rank`
from Scores180. Consecutive Numbers
Difficulty: Medium
SQL Schema
Write an SQL query to find all numbers that appear at least three times consecutively.
+----+-----+
| Id | Num |
+----+-----+
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 2 |
| 5 | 1 |
| 6 | 2 |
| 7 | 2 |
+----+-----+For example, given the above Logs table, 1 is the only number that appears consecutively for at least three times.
+-----------------+
| ConsecutiveNums |
+-----------------+
| 1 |
+-----------------+select distinct Num ConsecutiveNums
from
(
select
Num,
lead(Num,1,null) over(order by id) n2,
lead(Num,2,null) over(order by id) n3
from Logs
)t1
where Num = n2 and Num = n3181. Employees Earning More Than Their Managers
Difficulty: Easy
SQL Schema
The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.
+----+-------+--------+-----------+
| Id | Name | Salary | ManagerId |
+----+-------+--------+-----------+
| 1 | Joe | 70000 | 3 |
| 2 | Henry | 80000 | 4 |
| 3 | Sam | 60000 | NULL |
| 4 | Max | 90000 | NULL |
+----+-------+--------+-----------+Given the Employee table, write a SQL query that finds out employees who earn more than their managers. For the above table, Joe is the only employee who earns more than his manager.
+----------+
| Employee |
+----------+
| Joe |
+----------+select a.Name Employee
from Employee a
join Employee b
on a.ManagerId = b.id
where a.Salary>b.Salary182. Duplicate Emails
Difficulty: Easy
SQL Schema
Write a SQL query to find all duplicate emails in a table named Person.
Example:
+----+---------+
| Id | Email |
+----+---------+
| 1 | a@b.com |
| 2 | c@d.com |
| 3 | a@b.com |
+----+---------+Based on the above input, your query should return the following results:
+---------+
| Email |
+---------+
| a@b.com |
+---------+Note: All emails are in lowercase.
select Email
from Person
group by Email
having count(*)>1183. Customers Who Never Order
Difficulty: Easy
SQL Schema
Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL query to find all customers who never order anything.
Customers table:
+----+-------+
| Id | Name |
+----+-------+
| 1 | Joe |
| 2 | Henry |
| 3 | Sam |
| 4 | Max |
+----+-------+Orders table:
+----+------------+
| Id | CustomerId |
+----+------------+
| 1 | 3 |
| 2 | 1 |
+----+------------+Using the above tables as example, return the following:
+-----------+
| Customers |
+-----------+
| Henry |
| Max |
+-----------+select c.Name Customers
from Customers c left join Orders o
on c.id = o.CustomerId
where o.id is null184. Department Highest Salary
Difficulty: Medium
SQL Schema
The Employee table holds all employees. Every employee has an Id, a salary, and there is also a column for the department Id.
+----+-------+--------+--------------+
| Id | Name | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 70000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
+----+-------+--------+--------------+The Department table holds all departments of the company.
+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+Write a SQL query to find employees who have the highest salary in each of the departments. For the above tables, Max has the highest salary in the IT department and Henry has the highest salary in the Sales department.
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| Sales | Henry | 80000 |
+------------+----------+--------+select Department,Employee,Salary
from (
select d.Name Department,e.Name Employee, e.Salary,
rank() over(partition by d.id order by Salary desc) rk
from Employee e join Department d
on e.DepartmentId=d.id
)tmp
where rk = 1185. Department Top Three Salaries
Difficulty: Hard
SQL Schema
The Employee table holds all employees. Every employee has an Id, and there is also a column for the department Id.
+----+-------+--------+--------------+
| Id | Name | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 85000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
| 5 | Janet | 69000 | 1 |
| 6 | Randy | 85000 | 1 |
| 7 | Will | 70000 | 1 |
+----+-------+--------+--------------+The Department table holds all departments of the company.
+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+Write a SQL query to find employees who earn the top three salaries in each of the departments. For the above tables, your query should return the following rows:
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| IT | Randy | 85000 |
| IT | Joe | 85000 |
| IT | Will | 70000 |
| Sales | Henry | 80000 |
| Sales | Sam | 60000 |
+------------+----------+--------+Explanation:
In the IT department:
- Max earns the highest unique salary
- Both Randy and Joe earn the second-highest unique salary
- Will earns the third-highest unique salary
In the Sales department:
- Henry earns the highest salary
- Sam earns the second-highest salary
select Department,Employee,Salary
from (
select d.Name Department,e.Name Employee, e.Salary,
dense_rank() over(partition by d.id order by Salary desc) rk
from Employee e join Department d
on e.DepartmentId=d.id
)tmp
where rk <=3196. Delete Duplicate Emails
Difficulty: Easy
Write a SQL query to delete all duplicate email entries in a table named Person, keeping only unique emails based on its smallest Id.
+----+------------------+
| Id | Email |
+----+------------------+
| 1 | john@example.com |
| 2 | bob@example.com |
| 3 | john@example.com |
+----+------------------+
Id is the primary key for this table.For example, after running your query, the above Person table should have the following rows:
+----+------------------+
| Id | Email |
+----+------------------+
| 1 | john@example.com |
| 2 | bob@example.com |
+----+------------------+Note:
- Your output is the whole
Persontable after executing your SQL. - Use the
deletestatement.
DELETE p1 FROM Person p1,
Person p2
WHERE
p1.Email = p2.Email AND p1.Id > p2.IdNote that this is a delete operation, not a query.
197. Rising Temperature
Difficulty: Easy
SQL Schema
Given a Weather table, write a SQL query to find all dates' Ids with higher temperature compared to its previous (yesterday's) dates.
+---------+------------------+------------------+
| Id(INT) | RecordDate(DATE) | Temperature(INT) |
+---------+------------------+------------------+
| 1 | 2015-01-01 | 10 |
| 2 | 2015-01-02 | 25 |
| 3 | 2015-01-03 | 20 |
| 4 | 2015-01-04 | 30 |
+---------+------------------+------------------+For example, return the following Ids for the above Weather table:
+----+
| Id |
+----+
| 2 |
| 4 |
+----+select
Id
from
(
select Id,RecordDate,Temperature,
lag(RecordDate,1,9999-99-99) over (order by RecordDate) yd,
lag(Temperature,1,999) over(order by RecordDate ) yt
from Weather
)tmp
where Temperature >yt
and datediff(RecordDate,yd)=1262. Trips and Users
Difficulty: Hard
SQL Schema
The Trips table holds all taxi trips. Each trip has a unique Id, while Client_Id and Driver_Id are both foreign keys to the Users_Id at the Users table. Status is an ENUM type of (‘completed’, ‘cancelled_by_driver’, ‘cancelled_by_client’).
+----+-----------+-----------+---------+--------------------+----------+
| Id | Client_Id | Driver_Id | City_Id | Status |Request_at|
+----+-----------+-----------+---------+--------------------+----------+
| 1 | 1 | 10 | 1 | completed |2013-10-01|
| 2 | 2 | 11 | 1 | cancelled_by_driver|2013-10-01|
| 3 | 3 | 12 | 6 | completed |2013-10-01|
| 4 | 4 | 13 | 6 | cancelled_by_client|2013-10-01|
| 5 | 1 | 10 | 1 | completed |2013-10-02|
| 6 | 2 | 11 | 6 | completed |2013-10-02|
| 7 | 3 | 12 | 6 | completed |2013-10-02|
| 8 | 2 | 12 | 12 | completed |2013-10-03|
| 9 | 3 | 10 | 12 | completed |2013-10-03|
| 10 | 4 | 13 | 12 | cancelled_by_driver|2013-10-03|
+----+-----------+-----------+---------+--------------------+----------+The Users table holds all users. Each user has an unique Users_Id, and Role is an ENUM type of (‘client’, ‘driver’, ‘partner’).
+----------+--------+--------+
| Users_Id | Banned | Role |
+----------+--------+--------+
| 1 | No | client |
| 2 | Yes | client |
| 3 | No | client |
| 4 | No | client |
| 10 | No | driver |
| 11 | No | driver |
| 12 | No | driver |
| 13 | No | driver |
+----------+--------+--------+Write a SQL query to find the cancellation rate of requests with unbanned users (both client and driver must not be banned) between Oct 1, 2013 and Oct 3, 2013. The cancellation rate is computed by dividing the number of canceled (by client or driver) requests with unbanned users by the total number of requests with unbanned users.
+------------+-------------------+
| Day | Cancellation Rate |
+------------+-------------------+
| 2013-10-01 | 0.33 |
| 2013-10-02 | 0.00 |
| 2013-10-03 | 0.50 |
+------------+-------------------+SELECT T.request_at AS `Day`,
ROUND(
SUM(
IF(T.STATUS = 'completed',0,1)
)
/
COUNT(T.STATUS),
2
) AS `Cancellation Rate`
FROM trips AS T
WHERE
T.Client_Id NOT IN (
SELECT users_id
FROM users
WHERE banned = 'Yes'
)
AND
T.Driver_Id NOT IN (
SELECT users_id
FROM users
WHERE banned = 'Yes'
)
AND T.request_at BETWEEN '2013-10-01' AND '2013-10-03'
GROUP BY T.request_at511. Game Play Analysis I
Difficulty: Easy
SQL Schema
Activity table:
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| player_id | int |
| device_id | int |
| event_date | date |
| games_played | int |
+--------------+---------+
The primary key for this table is (player_id, event_date).
This table shows the activity of players of some games.
Each row is a record of a player who logged in and played a number of games (possibly 0) before logging out on some day using some device.Write an SQL query to report the first login date for each player.
The query result format is in the following example:
Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-05-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result table:
+-----------+-------------+
| player_id | first_login |
+-----------+-------------+
| 1 | 2016-03-01 |
| 2 | 2017-06-25 |
| 3 | 2016-03-02 |
+-----------+-------------+select player_id ,event_date first_login
from (
select player_id ,event_date,
rank() over(partition by player_id order by event_date) rk
from Activity
) tmp
where rk = 1- Optimal (select the minimum date)
select player_id ,min(event_date) first_login
from Activity
group by player_id 512. Game Play Analysis II
Difficulty: Easy
SQL Schema
Table: Activity
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| player_id | int |
| device_id | int |
| event_date | date |
| games_played | int |
+--------------+---------+
(player_id, event_date) is the primary key of this table.
This table shows the activity of players of some games.
Each row is a record of a player who logged in and played a number of games (possibly 0) before logging out on some day using some device.Write a SQL query to report the device that is first logged in for each player.
The query result format is in the following example:
Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-05-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result table:
+-----------+-----------+
| player_id | device_id |
+-----------+-----------+
| 1 | 2 |
| 2 | 3 |
| 3 | 1 |
+-----------+-----------+select player_id ,device_id
from (
select player_id ,event_date,device_id,
rank() over(partition by player_id order by event_date) rk
from Activity
) tmp
where rk = 1534. Game Play Analysis III
Difficulty: Medium
SQL Schema
Table: Activity
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| player_id | int |
| device_id | int |
| event_date | date |
| games_played | int |
+--------------+---------+
(player_id, event_date) is the primary key of this table.
This table shows the activity of players of some games.
Each row is a record of a player who logged in and played a number of games (possibly 0) before logging out on some day using some device.Write an SQL query that reports for each player and date, how many games played so far by the player. That is, the total number of games played by the player until that date. Check the example for clarity.
The query result format is in the following example:
Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-05-02 | 6 |
| 1 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result table:
+-----------+------------+---------------------+
| player_id | event_date | games_played_so_far |
+-----------+------------+---------------------+
| 1 | 2016-03-01 | 5 |
| 1 | 2016-05-02 | 11 |
| 1 | 2017-06-25 | 12 |
| 3 | 2016-03-02 | 0 |
| 3 | 2018-07-03 | 5 |
+-----------+------------+---------------------+
For the player with id 1, 5 + 6 = 11 games played by 2016-05-02, and 5 + 6 + 1 = 12 games played by 2017-06-25.
For the player with id 3, 0 + 5 = 5 games played by 2018-07-03.
Note that for each player, we only care about the login date.Window function
select player_id,event_date ,
sum(games_played) over(partition by player_id order by event_date )games_played_so_far
from ActivitySelf-join
select
a1.player_id,
a1.event_date,
sum(a2.games_played) games_played_so_far
from Activity a1,Activity a2
where a1.player_id=a2.player_id and
a1.event_date>=a2.event_date
group by 1,2;550. Game Play Analysis IV
Difficulty: Medium
SQL Schema
Table: Activity
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| player_id | int |
| device_id | int |
| event_date | date |
| games_played | int |
+--------------+---------+
(player_id, event_date) is the primary key of this table.
This table shows the activity of players of some games.
Each row is a record of a player who logged in and played a number of games (possibly 0) before logging out on some day using some device.Write an SQL query that reports the fraction of players that logged in again on the day after the day they first logged in, rounded to 2 decimal places. In other words, you need to count the number of players that logged in for at least two consecutive days starting from their first login date, then divide that number by the total number of players.
The query result format is in the following example:
Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-03-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result table:
+-----------+
| fraction |
+-----------+
| 0.33 |
+-----------+
Only the player with id 1 logged back in after the first day he had logged in, so the answer is 1/3 = 0.33select round(avg(a.event_date is not null), 2) fraction
from
(select player_id, min(event_date) as login
from activity
group by player_id) p
left join activity a
on p.player_id=a.player_id and datediff(a.event_date, p.login)=1The use of
avghere is clever.After the
is not nullcheck, rows with an event_date value return 1, and nulls return 0.avgeffectively sums these (i.e., the count of ids meeting the condition) and divides by the total number of ids, giving the required fraction.
569. Median Employee Salary
Difficulty: Hard
SQL Schema
The Employee table holds all employees. The employee table has three columns: Employee Id, Company Name, and Salary.
+-----+------------+--------+
|Id | Company | Salary |
+-----+------------+--------+
|1 | A | 2341 |
|2 | A | 341 |
|3 | A | 15 |
|4 | A | 15314 |
|5 | A | 451 |
|6 | A | 513 |
|7 | B | 15 |
|8 | B | 13 |
|9 | B | 1154 |
|10 | B | 1345 |
|11 | B | 1221 |
|12 | B | 234 |
|13 | C | 2345 |
|14 | C | 2645 |
|15 | C | 2645 |
|16 | C | 2652 |
|17 | C | 65 |
+-----+------------+--------+Write a SQL query to find the median salary of each company. Bonus points if you can solve it without using any built-in SQL functions.
+-----+------------+--------+
|Id | Company | Salary |
+-----+------------+--------+
|5 | A | 451 |
|6 | A | 513 |
|12 | B | 234 |
|9 | B | 1154 |
|14 | C | 2645 |
+-----+------------+--------+select Id,Company,Salary
from (
select Id,Company,Salary,
ROW_NUMBER() over(partition by Company order by Salary) rk,
count(*) over(partition by Company) cnt
from Employee
)t1
where rk IN (FLOOR((cnt + 1)/2), FLOOR((cnt + 2)/2))Median:
+1 floor and +2 floor
570. Managers with at Least 5 Direct Reports
Difficulty: Medium
SQL Schema
The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.
+------+----------+-----------+----------+
|Id |Name |Department |ManagerId |
+------+----------+-----------+----------+
|101 |John |A |null |
|102 |Dan |A |101 |
|103 |James |A |101 |
|104 |Amy |A |101 |
|105 |Anne |A |101 |
|106 |Ron |B |101 |
+------+----------+-----------+----------+Given the Employee table, write a SQL query that finds out managers with at least 5 direct report. For the above table, your SQL query should return:
+-------+
| Name |
+-------+
| John |
+-------+Note: No one would report to themselves.
select Name
from Employee
where Id in (
select ManagerId
from Employee
group by ManagerId
having count(*)>=5
)571. Find Median Given Frequency of Numbers
Difficulty: Hard
SQL Schema
The Numbers table keeps the value of number and its frequency.
+----------+-------------+
| Number | Frequency |
+----------+-------------|
| 0 | 7 |
| 1 | 1 |
| 2 | 3 |
| 3 | 1 |
+----------+-------------+In this table, the numbers are 0, 0, 0, 0, 0, 0, 0, 1, 2, 2, 2, 3, so the median is (0 + 0) / 2 = 0.
+--------+
| median |
+--------|
| 0.0000 |
+--------+Write a query to find the median of all numbers and name the result as median.
select
avg(cast(number as float)) median
from
(
select
Number,
Frequency,
sum(Frequency) over(order by Number) - Frequency prev_sum,
sum(Frequency) over(order by Number) curr_sum
from Numbers
) t1,
(
select
sum(Frequency) total_sum
from Numbers
) t2
where
t1.prev_sum <= (cast(t2.total_sum as float) / 2) and
t1.curr_sum >= (cast(t2.total_sum as float) / 2)If n1.Number is the median, the cumulative count of numbers before n1.Number (including itself) should be greater than or equal to total/2, and the cumulative count before n1.Number (excluding itself) should be less than or equal to total/2.
For example: 0, 0, 0, 0, 0, 0, 0, 1, 2, 2, 2, 3 (12 numbers total)
Median 0 (including itself) cumulative count is 7 >= 6; 0 (excluding itself) cumulative count is 0 <= 6. For example: 0, 0, 0, 3, 3, 3 (6 numbers total)
Median 0 (including itself) cumulative count is 3 >= 3; 0 (excluding itself) cumulative count is 0 <= 3.
Median 3 (including itself) cumulative count is 6 >= 3; 3 (excluding itself) cumulative count is 3 <= 3.
SELECT
AVG(Number)median
FROM
(SELECT n1.Number FROM Numbers n1 JOIN Numbers n2 ON n1.Number>=n2.Number
GROUP BY
n1.Number
HAVING
SUM(n2.Frequency)>=(SELECT SUM(Frequency) FROM Numbers)/2
AND
SUM(n2.Frequency)-AVG(n1.Frequency)<=(SELECT SUM(Frequency) FROM Numbers)/2
)s574. Winning Candidate
Difficulty: Medium
SQL Schema
Table: Candidate
+-----+---------+
| id | Name |
+-----+---------+
| 1 | A |
| 2 | B |
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